In this paper we present some trace inequalities for positive definite matrices in statistical mechanics

ON SOME MATRIX INEQUALITIES Hyun Deok Lee

Abstract. In this paper we present some trace inequalities for positive definite matrices in statistical mechanics. In order to prove the method of the uniform bound on the generating functional for the semi-classical model, we use some trace inequalities and matrix norms and properties of trace for positive definite matrices.

1. Introduction

Matrix inequalities play an important role in statistical mechan- ics([1,3,6,7]). We study quantum statistical mechanics for the semi- classical model in the lattice space. In order to investigate the uniform bound on the generating functional, the semi-classical model has been studied intensively by many authors([1,3,6,7,10]). The purpose of this paper is to establish the trace inequality for multiple product of pow- ers of arbitrary finite positive definite matrices which is a tool to find the uniform bound on the generating functional for the semi-classical model in quantum statistical mechanics.

This paper is organized as follows. In section II, We introduce some definitions and theorems which are necessary to prove our main result and present some properties of trace for positive definite matrices on the finite dimensional Hilbert space([ 2,4,11 ]). In section III, we first describe some trace inequalities, and then we will prove our main result by applying these trace inequalities and properties of matrix norms,and induction argument.

2. Preliminaries

Let Hn denote the complex vector space of all n × n Hermitian

Received November 20, 2008. Revised December 7, 2008.

2000 Mathematics Subject Classification: 82B10.

Key words and phrases: positive definite matrices, matrix norms, trace.

This work was supported by the Cheongju University Scholar Overseas Funds.

matrices, endowed with the inner product < A, B >= T r(B^∗A), where T r(·) is the trace on the positive matrices and B^∗ is the adjoint of B, Then,this makes (Hn, < ·, · >) into a Hilbert space H([11,12]).

We first define an n × n Hermitian matrix A is said to be positive definite, denoted by, A > 0, if x^∗Ax > 0 for all nonzero x in Cⁿ. If x^∗Ax ≥ 0,then A is said to be positive semidefinite, denoted by, A ≥ 0.

Definition 2.1. ([10,12]) Let A in Hn, Then, (1)The trace norm of A, defined by, kAk1 = P

isi(A). (2) The spectral norm of A,also denoted by, kAk2 = max {si(A)}, where si(A) are the singular values of A, i.e., the eigenvalues of |A| = (A^∗A)¹².

Theorem 2.2. ([6])Let A be a positive definite matrix in H_n with finite trace norm and B be a positive matrix with finite trace norm, Then,

(1)T r(U AU⁻¹) = T r(A) for any unitary matrix U ,

(2)T r(AB) = T r(BA).

We now restrict to the case the finite dimension of H, i.e., dim(H) < ∞

Theorem 2.3. ([11])[Cauchy-Schwartz inequality] Let A and B be positive definite matrices with finite trace norm,respectively, then,

|T r(A^∗B)|² ≤ T r(A^∗A)T r(B^∗B).

By the Cauchy-Schwartz inequality, we have

Theorem 2.4. ([11]) Let A and B be positive definite matrices with finite trace norm, respectively, then,

(1)T r(AB) ≤ |T r(AB)| ≤ {T r(A^∗A)}¹²{T r(B^∗B)}¹²,

(2)T r(AB) < T r(A)T r(B).

3. Trace Inequalities

In order to prove our main result, it is necessary to establish the following two lemmas.

Lemma 3.1. ([4,5,9])Let A ≥ 0 and B ≥ 0 in Hn, then, T r(AB) ≤ kAk2T r(B),

where kAk₂ denotes the spectral norm or largest singular value of A.

Lemma 3.2. ([9,11 ])For any positive matrices C,D, and E in H_n, then,

|T r(CDE)| ≤ kDk2{T r(C^∗C)}¹²{T r(E^∗E)}¹². Proof.

|T r(CDE)| ≤ kDk2T r(CE)

By Theorem 2.2 and Theorem 2.4, combining the Cauchy-Schwartz inequality, we obtain the following :

|T r(CDE)| ≤ kDk2T r(CE) ≤ kDk2{T r(C^∗C)}¹²{T r(E^∗E)}¹².

¤ Now, we will prove our main result using above two lemmas.

Theorem 3.3. Let B1, B2, ... , Bn be positive definite matrices with finite matrix norm in H_n, and A be a positive definite matrix such that Aⁿ¹ has a finite trace norm, then, for pi ≥ 0, i = 1, 2, ..., n, with P_n

i=1p_i = 1,

|T r(B₁A^p1B₂A^p2 · · · B_nA^pn)| ≤ ( Yn i=1

kB_ik₂)T r(A).

Proof. We will prove the theorem by induction. For n = 1, it follows from Lemma 3.1. For n = 2, let p1 ≤ ¹₂,then,

|T r(B1A^p1B2A^p2)|

= |T r(A^p1+p22 B1A^p1B2A^p1−p22 )|

≤ kB1k2T r(A)¹²T r(A^p1−p22 B₂^∗A^2p1B2A^p1−p22 )¹², by Lemma 3.2.

Continuing in this process, after n-steps,we obtain

|T r(B1A^p1B2A^p2)|

≤ kB1k2kB2k2

1

2+···+_2n−1¹ T r(A)^{12+···+2n−11} |T r(B₂^∗A^p10B2A^p20)|

1 2n

for some p1

0 ≥ 0 , p2

0 ≥ 0 with p1

0 + p2

0 = 1.

Since one of pi

0’s is less than ¹₂, we have

|T r(B^∗₂A^p10B2A^p20)|

1

2n ≤ kB2k2

1 2nkAk2

1

2nT r(A¹²)

1 2n. By taking n −→ ∞, we proved the theorem for n = 2.

We now assume that the theorem holds for n ≤ m − 1. We will show that the theorem holds for n = m.

Since p1+ p2+ ... + pm = 1, there exists j in N such that pj + pj+1 (mod m)+ ... + p_j+[^m_{2] (mod m)} < 1

2 pj+ pj+1 (mod m) + ... + p_j+1+[^m_{2] (mod m)}≥ 1

2, where [^m₂] is the largest integer which is not greater than ^m₂.

Using the cyclic property of the trace : T r(AB) = T r(BA) and rearranging Bj and pj, we may assume that

p1+ p2+ ... + p_[^m_2] < 1 2 p1+ p2+ ... + p_[^m_2]+1 ≥ 1

2 · · · ·(1) Let p⁰ = ¹₂ −P_[^m_2]

i=1pi and let m⁰ = [^m₂] + 1, then,

2(

mX⁰−1 i=1

pi+ p⁰) = 1

2(

Xm i=m⁰+1

p_i+ (p_m⁰ − p⁰)) = 1 · · · ·(2)

and

|T r(B1A^p1B2A^p2· · · BmA^pm)|

≤ |T r(A^p0m−p0B_m⁰₊₁A^pm0+1 · · · B_mA^pmB₁A^p1 · · · B_m⁰A^p0)|

≤ kB₁k₂T r(A^p0B_m^∗0A^pm0−1B_m⁰₋₁...B₂^∗A^2p1B₂A^p2...B_m⁰A^p10)¹²

T r(A^pm0−p0B_m⁰₊₁...B_m^∗A^2pmB_mA^pm−1...B_m⁰₊₁A^pm0−p0)¹² · · · ·(3) If m is odd, then 2(m⁰− 1) = m − 1 and 2(m − (m⁰+ 1)) = m − 1.

Thus, for odd integer m, the theorem follows from (2) and the induction hypothesis.

Next, let m be even, then 2(m⁰−1) = m and 2(m−(m⁰+1)) = m−2.

So, by the induction hypothesis, (3) is bounded by

kB₁k₂ Ym j=m⁰+1

T r(A)¹²T r(B_m^∗0A^pm0−1B_m−1^∗ ...B₂^∗A^2p1B₂...B_m⁰A^2p0)¹²···(4)

Notice that by (1) either

mX⁰−1 j=2

pj < 1 2 and

mX⁰−1 j=2

p_j + 2p⁰ ≥ 1

2 · · · ·(5) or else

mX⁰−1 j=2

pj < 1 2 and

mX⁰−1 j=2

pj+ 2p1 ≥ 1

2 · · · ·(6)

In either case, we use the method to obtain (3) and (4).

T r(B_m^∗0A^pm0−1B_m^∗0₋₁· · · B2∗A^2p1B2A^p2...B_m⁰A^2p0)

≤ (

m⁰

Y

j=2

kBjk2T r(A)¹²T r(A^pm0B_m^∗0 · · · B₂^∗A^2p10B2A^p20 · · · B_m⁰A^pm0)

for some p1

0, ..., pm

0 with 2(p1

0 + ... + pm

0) = 1 and such that one of (5) and (6) holds for p1

0, ..., pm

0

After n-steps of the above continuing process, we conclude that

|T r(B1A^p1...BmA^pm)|

≤ kB₁k₂( Ym j=m⁰+1

kB_jk₂)T r(A)^{12+···+2n−11} (

m⁰

Y

j=2

kB_jk₂)^{12+···+2n−11}

|T r(A^qm0B^∗_m⁰ · · · B₂^∗A^2q1B2· · · B_m⁰A^qm0)|²ⁿ¹ · · · ·(7) for some q₁, ..., q_m⁰ with 2(q₁+ ... + q_m⁰) = 1

Since one of qi’s is less than _m¹, the trace in (7) is bounded by (Q_m⁰

j=2kBjk

1

22n−1) kAk^(1−m1)2n1 T r(A^m1 )²ⁿ¹ .

Hence, The theorem follows from (7). This completes the proof of

the theorem. ¤

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Department of Mathematics Education Cheongju University

Cheongju 360-764, Korea E-mail: hdlee@cju.ac.kr